CBSE Class 10 Mathematics Quadratic Equations VBQs Set D

Know the Terms

• The real roots of \( ax^2 + bx + c = 0 \), where \( a \neq 0 \) are \( \frac{-b + \sqrt{b^2 - 4ac}}{2a} \) and \( \frac{-b - \sqrt{b^2 - 4ac}}{2a} \), where \( b^2 - 4ac > 0 \).
• Roots of \( ax^2 + bx + c = 0 \), where \( a \neq 0 \) are \( \frac{-b}{2a} \) and \( \frac{-b}{2a} \), where \( b^2 - 4ac = 0 \).
• Quadratic identities:
  • (i) \( (a + b)^2 = a^2 + 2ab + b^2 \)
  • (ii) \( (a - b)^2 = a^2 - 2ab + b^2 \)
  • (iii) \( a^2 - b^2 = (a + b)(a - b) \)
• Discriminant, \( D = b^2 - 4ac \).

(A) OBJECTIVE TYPE QUESTIONS

Stand Alone MCQs

Question. Which of the following is a quadratic equation?
(a) \( x^2 + 2x + 1 = (4 - x)^2 + 3 \)
(b) \( -2x^2 = (5 - x)(2x - \frac{2}{5}) \)
(c) \( (k + 1)x^2 + \frac{3}{2}x = 7 \), where \( k = -1 \)
(d) \( x^3 - x^2 = (x - 1)^3 \)
Answer: (d)
Explanation: \( x^3 - x^2 = x^3 - 1 - 3x(x - 1) \)
\( x^3 - x^2 = x^3 - 1 - 3x^2 + 3x \)
\( -x^2 + 3x^2 - 3x + 1 = 0 \)
\( 2x^2 - 3x + 1 = 0 \)
It is of the form \( ax^2 + bx + c = 0 \).

Question. Which of the following is not a quadratic equation?
(a) \( 2(x - 1)^2 = 4x^2 - 2x + 1 \)
(b) \( 2x - x^2 = x^2 + 5 \)
(c) \( (\sqrt{2}x + \sqrt{3})^2 + x^2 = 3x^2 - 5x \)
(d) \( (x^2 + 2x)^2 = x^4 + 3 + 4x^3 \)
Answer: (c)
Explanation: \( (\sqrt{2}x)^2 + (\sqrt{3})^2 + 2 \times \sqrt{2}x \times \sqrt{3} + x^2 = 3x^2 - 5x \)
\( 2x^2 + 3 + 2\sqrt{6}x + x^2 = 3x^2 - 5x \)
\( 3x^2 + 2\sqrt{6}x + 3 = 3x^2 - 5x \)
\( x(5 + 2\sqrt{6}) + 3 = 0 \)
It is not of the form of \( ax^2 + bx + c = 0 \).

Question. Which of the following equations has 2 as it's root?
(a) \( x^2 - 4x + 5 = 0 \)
(b) \( x^2 + 3x - 12 = 0 \)
(c) \( 2x^2 - 7x + 6 = 0 \)
(d) \( 3x^2 - 6x - 2 = 0 \)
Answer: (c)
Explanation: Put the value of \( x = 2 \) in \( 2x^2 - 7x + 6 = 0 \)
\( 2(2)^2 - 7(2) + 6 = 0 \)
\( 8 - 14 + 6 = 0 \)
\( 0 = 0 \)
So, \( x = 2 \) is a root of \( 2x^2 - 7x + 6 = 0 \).

Question. If \( \frac{1}{2} \) is a root of the equation \( x^2 + kx - \frac{5}{4} = 0 \), then the value of k is:
(a) 2
(b) -2
(c) \( \frac{1}{2} \)
(d) \( \frac{1}{2} \)
Answer: (a)
Explanation: Since, \( \frac{1}{2} \) is a root of the equation \( x^2 + kx - \frac{5}{4} = 0 \),
Then, \( (\frac{1}{2})^2 + k(\frac{1}{2}) - \frac{5}{4} = 0 \)
\( \frac{1}{4} + \frac{k}{2} - \frac{5}{4} = 0 \)
\( \frac{k}{2} = \frac{5}{4} - \frac{1}{4} \)
\( \frac{k}{2} = 1 \)
\( k = 2 \).

Question. Which of the following equations has the sum of its roots as 3?
(a) \( 2x^2 - 3x + 6 = 0 \)
(b) \( -x^2 + 3x - 3 = 0 \)
(c) \( \sqrt{2}x^2 - \frac{3}{\sqrt{2}}x + 1 = 0 \)
(d) \( 3x^2 - 3x + 3 = 0 \)
Answer: (b)
Explanation: \( -x^2 + 3x - 3 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \)
\( a = -1, b = 3, c = -3 \)
Sum of the roots \( = \frac{-b}{a} = \frac{-3}{-1} = 3 \)

Question. The roots of the equation \( x^2 + 7x + 10 = 0 \) are:
(a) -5, -2
(b) 5, 2
(c) 5, -2
(d) -5, 2
Answer: (a)
Explanation: Given, \( x^2 + 7x + 10 = 0 \)
Comparing with \( ax^2 + bx + c = 0 \), we get \( a = 1, b = 7 \) and \( c = 10 \)
\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
\( x = \frac{-7 \pm \sqrt{(7)^2 - 4 \times 1 \times 10}}{2 \times 1} \)
\( x = \frac{-7 \pm \sqrt{9}}{2} \)
\( x = \frac{-7 \pm 3}{2} \)
\( x = \frac{-7 + 3}{2} \) or \( \frac{-7 - 3}{2} \)
\( x = -2 \) or \(-5 \)
Hence, the roots of the given equation are -2 and -5.

Question. Which of the following equations has two distinct real roots?
(a) \( 2x^2 - 3\sqrt{2}x + \frac{9}{4} = 0 \)
(b) \( x^2 + x - 5 = 0 \)
(c) \( x^2 + 3x + 2\sqrt{2} = 0 \)
(d) \( 5x^2 - 3x + 1 = 0 \)
Answer: (b)
Explanation: \( x^2 + x - 5 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \)
\( a = 1, b = 1, c = -5 \)
\( b^2 - 4ac = (1)^2 - 4(1)(-5) \)
\( = 1 + 20 \)
\( = 21 > 0 \)
Since \( D \) (i.e., \( b^2 - 4ac > 0 \))
Hence, the equation has two distinct real roots.

Question. Values of \( k \) for which the quadratic equation \( 2x^2 - kx + k = 0 \) has equal roots, is:
(a) 0 only
(b) 4
(c) 8 only
(d) 0, 8
Answer: (d)
Explanation: Given equation is \( 2x^2 - kx + k = 0 \)
On comparing with \( ax^2 + bx + c = 0 \)
\( a = 2, b = -k, c = k \)
For equal roots \( b^2 - 4ac = 0 \)
\( (-k)^2 - 4(2)(k) = 0 \)
\( k^2 - 8k = 0 \)
\( k(k - 8) = 0 \)
\( k = 0, 8 \)
Hence, the required values of \( k \) are 0 and 8.

Question. The quadratic equation \( 2x^2 - \sqrt{5}x + 1 = 0 \) has
(a) two distinct real roots
(b) two equal real roots
(c) no real root
(d) more than 2 real roots
Answer: (c)
Explanation: \( 2x^2 - \sqrt{5}x + 1 = 0 \)
On comparing with \( ax^2 + bx + c = 0 \)
\( a = 2, b = -\sqrt{5}, c = 1 \)
Discriminant \( = b^2 - 4ac = (-\sqrt{5})^2 - 4(2)(1) \)
\( = 5 - 8 = -3 < 0 \)
Since \( D \) (i.e., \( b^2 - 4ac < 0 \))
Therefore, the equation has no real root.

Question. Which of the following equations has no real root?
(a) \( x^2 - 4x + 3\sqrt{2} = 0 \)
(b) \( x^2 + 4x - 3\sqrt{2} = 0 \)
(c) \( x^2 - 4x - 3\sqrt{2} = 0 \)
(d) \( 3x^2 + 4\sqrt{3}x + 4 = 0 \)
Answer: (a)
Explanation: \( x^2 - 4x + 3\sqrt{2} = 0 \)
On comparing with \( ax^2 + bx + c = 0 \)
\( a = 1, b = -4, c = 3\sqrt{2} \)
Discriminant \( = b^2 - 4ac \)
\( = (-4)^2 - 4(1)(3\sqrt{2}) \)
\( = 16 - 12\sqrt{2} \)
\( = 16 - 12 \times 1.41 \)
\( = 16 - 16.92 \)
\( = -0.92 < 0 \)
Since, \( D < 0 \)
Therefore, the equation has no real roots.

Question. \( (x^2 + 1)^2 - x^2 = 0 \) has
(a) four real roots
(b) two real roots
(c) no real roots
(d) one real root.
Answer: (c)
Explanation: \( (x^2 + 1)^2 - x^2 = 0 \)
\( x^4 + 1 + 2x^2 - x^2 = 0 \)
\( x^4 + x^2 + 1 = 0 \)
Let \( x^2 = y \)
\( y^2 + y + 1 = 0 \)
On comparing with \( ay^2 + by + c = 0 \)
\( a = 1, b = 1, c = 1 \)
Discriminant, \( D = b^2 - 4ac \)
\( = (1)^2 - 4(1)(1) \)
\( = 1 - 4 \)
\( = -3 < 0 \)
Since \( D < 0 \)
Therefore, the equation has no real roots.

Case-based MCQs

Raj and Ajay are very close friends. Both the families decide to go to Ranikhet by their own cars. Raj’s car travels at a speed of \( x \) km/h while Ajay’s car travels 5 km/h faster than Raj’s car. Raj took 4 hours more than Ajay to complete the journey of 400 km.

Question. What will be the distance covered by Ajay’s car in two hours?
(a) \( 2(x + 5) \) km
(b) \( (x - 5) \) km
(c) \( 2(x + 10) \) km
(d) \( (2x + 5) \) km
Answer: (a)
Explanation: Speed of Raj's car \( = x \) km/hr
Speed of Ajay's car \( = (x + 5) \) km/hr
Distance covered by Ajay in 2 hours \( = [(x + 5) \times 2] \) km \( = 2(x + 5) \) km.

Question. Which of the following quadratic equation describe the speed of Raj’s car?
(a) \( x^2 - 5x - 500 = 0 \)
(b) \( x^2 + 4x - 400 = 0 \)
(c) \( x^2 + 5x - 500 = 0 \)
(d) \( x^2 - 4x + 400 = 0 \)
Answer: (c)
Explanation: Given,
Speed of Raj's car \( = x \) km/hour
Speed of Ajay's car \( = x + 5 \) km/hour
Time taken by Raj \( = \frac{400}{x} \) hours
Time taken by Ajay \( = \frac{400}{(x + 5)} \) hours
Raj took 4 hours more than Ajay i.e.,
\( \frac{400}{x} - \frac{400}{x + 5} = 4 \)
\( \Rightarrow \frac{100}{x} - \frac{100}{x + 5} = 1 \)
\( \Rightarrow 100(x + 5) - 100x = x(x + 5) \)
\( \Rightarrow x^2 + 5x - 500 = 0 \)

Question. What is the speed of Raj’s car?
(a) 20 km/hour
(b) 15 km/hour
(c) 25 km/hour
(d) 10 km/hour
Answer: (a)
Explanation: Solving quadratic equation \( x^2 + 5x - 500 = 0 \), we get
\( (x + 25)(x - 20) = 0 \)
\( \Rightarrow x = -25 \) or \( x = 20 \)
\( \Rightarrow x = 20 \) (as \( x \) can't be negative)
So, speed of Raj's car is 20 km/hour.

Question. How much time took Ajay to travel 400 km?
(a) 20 h
(b) 40 h
(c) 25 h
(d) 16 h
Answer: (d)
Explanation: Time taken by Ajay \( = \frac{400}{x + 5} \) h
\( = \frac{400}{20 + 5} \) h
\( = 16 \) h

Question. The Speed of Ajay’s car is:
(a) 25 km/hour
(b) 50 km/hour
(c) 75 km/hour
(d) None of these
Answer: (a)
Explanation: Speed of Ajay's car \( = x + 5 \) km/hour
\( = 20 + 5 \)
\( = 25 \) km/hour

The speed of a motor boat is 20 km/hr., For covering the distance of 15 km the boat took 1 hour more for upstream than downstream.

Question. Let speed of the stream be \( x \) km/hr. then speed of the motorboat in upstream will be
(a) 20 km/hr
(b) \( (20 + x) \) km/hr
(c) \( (20 - x) \) km/hr
(d) 2 km/hr
Answer: (c)
Explanation: Speed of motorboat in upstream \( = \) Speed of motorboat \( - \) Speed of stream \( = (20 - x) \) km/hr.

Question. What is the relation between speed, distance and time?
(a) speed \( = \) (distance)/time
(b) distance \( = \) (speed)/time
(c) time \( = \) speed \( \times \) distance
(d) speed \( = \) distance \( \times \) time
Answer: (a)

Question. Which is the correct quadratic equation for the speed of the current?
(a) \( x^2 + 30x - 200 = 0 \)
(b) \( x^2 + 20x - 400 = 0 \)
(c) \( x^2 + 30x - 400 = 0 \)
(d) \( x^2 - 20x - 400 = 0 \)
Answer: (c)
Explanation: In downstream:
Speed \( = (20 + x) \) km/hour
Distance \( = 15 \) km
Time \( = \frac{15}{20 + x} \)
In upstream:
Speed \( = (20 - x) \) km/hour
Distance \( = 15 \) km
Time \( = \frac{15}{20 - x} \)
According to question, for covering the distance of 15 km the boat took 1 hour more for upstream than downstream.
So, \( \frac{15}{20 - x} - \frac{15}{20 + x} = 1 \)
\( \Rightarrow \frac{15(20 + x) - 15(20 - x)}{(20 - x)(20 + x)} = 1 \)
\( \Rightarrow 30x = 400 - x^2 \)
\( \Rightarrow x^2 + 30x - 400 = 0 \)

Question. What is the speed of current?
(a) 20 km/hour
(b) 10 km/hour
(c) 15 km/hour
(d) 25 km/hour
Answer: (b)
Explanation: On solving the quadratic equation \( x^2 + 30x - 400 = 0 \), we get
\( (x - 10)(x + 40) = 0 \)
\( \Rightarrow x = 10 \) or \( x = -40 \)
So, Speed of the stream \( = 10 \) km/hour

Question. How much time boat took in downstream?
(a) 90 minute
(b) 15 minute
(c) 30 minute
(d) 45 minute
Answer: (c)
Explanation: Time taken in downstream \( = \frac{15}{20 + x} = \frac{15}{20 + 10} = \frac{1}{2} \) hours \( = 30 \) minutes.

John and Jivanti are playing with the marbles. They together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they now have is 124.

Question. If John had \( x \) number of marbles, then number of marbles Jivanti had:
(a) \( x - 45 \)
(b) \( 45 - x \)
(c) \( 45x \)
(d) \( x - 5 \)
Answer: (b)
Explanation: If John had \( x \) number of marbles, then Jivanti had \( (45 - x) \) marbles, because there are total 45 marbles.

Question. Number of marbles left with Jivanti, when she lost 5 marbles:
(a) \( x - 45 \)
(b) \( 40 - x \)
(c) \( 45 - x \)
(d) \( x - 40 \)
Answer: (b)
Explanation: Number of marbles left with Jivanti, when she lost 5 marbles \( = (45 - x - 5) = (40 - x) \)

Question. The quadratic equation related to the given problem is:
(a) \( x^2 - 45x + 324 = 0 \)
(b) \( x^2 + 45x + 324 = 0 \)
(c) \( x^2 - 45x - 324 = 0 \)
(d) \( -x^2 - 45x + 324 = 0 \)
Answer: (a)
Explanation: According to question,
\( (x - 5)(40 - x) = 124 \)
\( \Rightarrow -x^2 + 200 + 40x + 5x - 124 = 0 \)
\( \Rightarrow x^2 - 45x + 324 = 0 \)

Question. Number of marbles John had:
(a) 10
(b) 9
(c) 35
(d) 30
Answer: (b)
Explanation: \( x^2 - 45x + 324 = 0 \)
\( \Rightarrow x^2 - 9x - 36x + 324 = 0 \)
\( \Rightarrow x(x - 9) - 36(x - 9) = 0 \)
\( \Rightarrow (x - 9)(x - 36) = 0 \)
Either \( x = 9 \) or \( x = 36 \).
Therefore, the number of marbles John had 9 or 36.

Question. If John had 36 marbles, then number of marbles Jivanti had:
(a) 10
(b) 9
(c) 36
(d) 35
Answer: (b)
Explanation: If John had 36 marbles, then Jivanti had \( (45 - 36) = 9 \) marbles.

There is a triangular playground as shown in the figure below. Many Children and people are playing and walking in the ground. As we see in the above figure of right angled triangle playground, the length of the sides are \( 5x \) cm and \( (3x - 1) \) cm and area of the triangle is \( 60 \text{ cm}^2 \).

Question. The value of \( x \) is:
(a) 8
(b) 3
(c) 4
(d) 5
Answer: (b)
Explanation: Given, area of triangle \( = 60 \text{ cm}^2 \)
\( \Rightarrow \frac{1}{2} \times AB \times BC = 60 \)
\( \Rightarrow (5x)(3x - 1) = 120 \)
\( \Rightarrow 3x^2 - x - 24 = 0 \)
\( \Rightarrow 3x^2 - 9x + 8x - 24 = 0 \)
\( \Rightarrow 3x(x - 3) + 8(x - 3) = 0 \)
\( \Rightarrow (x - 3)(3x + 8) = 0 \)
Either \( x = 3 \) or \( x = -\frac{8}{3} \)
Since length can't be negative, then \( x = 3 \).

Question. The length of AB is:
(a) 8 cm
(b) 10 cm
(c) 15 cm
(d) 17 cm
Answer: (c)
Explanation: The length of \( AB = 5x \text{ cm} = 5 \times 3 \text{ cm} = 15 \text{ cm} \).

Question. The length of AC is:
(a) 17 cm
(b) 15 cm
(c) 21 cm
(d) 20 cm
Answer: (a)
Explanation: \( AB = 15 \text{ cm} \) and \( BC = (3x - 1) \text{ cm} = (3 \times 3 - 1) \text{ cm} = 8 \text{ cm} \)
Now, in right angled \( \Delta ABC \),
\( AC^2 = AB^2 + BC^2 \) (By using Pythagoras theorem)
\( = (15)^2 + (8)^2 \)
\( = 225 + 64 \)
\( = 289 = (17)^2 \)
Hence, \( AC = 17 \text{ cm} \).

Question. The perimeter of \( \Delta ABC \) is:
(a) 35 cm
(b) 45 cm
(c) 30 cm
(d) 40 cm
Answer: (d)
Explanation: Here, \( AB = 15 \text{ cm}, BC = 8 \text{ cm} \) and \( AC = 17 \text{ cm} \).
Then, the perimeter of \( \Delta ABC = (AB + BC + CA) \text{ cm} \)
\( = (15 + 8 + 17) \text{ cm} = 40 \text{ cm} \).

Question. The given problem is based on which mathematical concept?
(a) AP
(b) Linear equation in one variable
(c) Quadratic Equations
(d) None of these
Answer: (c)
Explanation: The given problem is based on the concept of quadratic equations.

(B) SUBJECTIVE QUESTIONS

Very Short Answer Type Questions

Question. Find the roots of the equation \( x^2 + 7x + 10 = 0 \).
Answer: \( x^2 + 7x + 10 = 0 \)
\( x^2 + 5x + 2x + 10 = 0 \)
\( (x + 5)(x + 2) = 0 \)
Either \( x = -5 \) and \( x = -2 \).

Question. Find the roots of the quadratic equation \( x^2 - 0.04 = 0 \).
Answer: \( x^2 - 0.04 = 0 \)
\( x^2 = 0.04 \)
\( x = \pm\sqrt{0.04} \)
\( x = \pm 0.2 \).

Question. If one root of the equation \( (k - 1)x^2 - 10x + 3 = 0 \) is the reciprocal of the other, then find the value of \( k \).
Answer: Let one root \( = \alpha \)
and the other root \( = \frac{1}{\alpha} \)
Product of roots \( = \alpha \times \frac{1}{\alpha} \)
Given equation is \( (k - 1)x^2 - 10x + 3 = 0 \)
Comparing it with \( ax^2 + bx + c = 0 \), we get \( a = (k - 1), b = 10 \) and \( c = 3 \)
Product of roots \( = \frac{c}{a} \)
\( \frac{3}{k - 1} = \alpha \times \frac{1}{\alpha} \)
\( \frac{3}{k - 1} = 1 \)
\( 3 = k - 1 \)
\( k = 4 \).

Question. Find the value of \( k \) for which the roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
Answer: Roots of the equation \( 3x^2 - 10x + k = 0 \) are reciprocal of each other.
\( \Rightarrow \) Product of roots \( = 1 \)
\( \Rightarrow \frac{k}{3} = 1 \Rightarrow k = 3 \).

Question. Write the discriminant of the quadratic equation \( (x + 5)^2 = 2(5x - 3) \).
Answer: \( (x + 5)^2 = 2(5x - 3) \)
\( x^2 + 25 + 10x = 10x - 6 \)
\( x^2 + 31 = 0 \)
\( a = 1, b = 0, c = 31 \)
Discriminant \( = b^2 - 4ac \)
\( = 0^2 - 4 \times 1 \times 31 \)
\( = 0 - 124 \)
\( = -124 \).

Question. If \( x = 3 \) is one root of the quadratic equation \( x^2 - 2kx - 6 = 0 \), then find the value of \( k \).
Answer: \( x = 3 \) is one root of the equation
\( \therefore 3^2 - 2k(3) - 6 = 0 \)
\( 9 - 6k - 6 = 0 \)
\( 6k = 3 \)
\( k = \frac{1}{2} \).

Question. Find the value of \( k \), for which one root of the quadratic equation \( kx^2 - 14x + 8 = 0 \) is 2.
Answer: Since, \( x = 2 \) is one root of the equation.
\( k(2)^2 - 14(2) + 8 = 0 \)
\( 4k - 28 + 8 = 0 \)
\( 4k = 20 \)
\( k = 5 \).

Question. Find the positive root of \( \sqrt{3x^2 + 6} = 9 \).
Answer: \( \sqrt{3x^2 + 6} = 9 \)
\( 3x^2 + 6 = 81 \)
\( 3x^2 = 81 - 6 = 75 \)
\( x^2 = \frac{75}{3} = 25 \)
\( x = \pm 5 \)
Hence, positive root \( = 5 \).

Question. If \( x = -\frac{1}{2} \), is a solution of the quadratic equation \( 3x^2 + 2kx - 3 = 0 \), find the value of \( k \).
Answer: Putting \( x = -\frac{1}{2} \) in \( 3x^2 + 2kx - 3 = 0 \)
\( 3(-\frac{1}{2})^2 + 2k(-\frac{1}{2}) - 3 = 0 \)
\( \Rightarrow \frac{3}{4} - k - 3 = 0 \)
\( k = \frac{3}{4} - 3 \)
\( k = \frac{3 - 12}{4} \)
\( k = \frac{-9}{4} \).

Question. Find the roots of the quadratic equation \( \sqrt{3}x^2 - 2x - \sqrt{3} = 0 \).
Answer: \( \sqrt{3}x^2 - 2x - \sqrt{3} = 0 \)
\( \sqrt{3}x^2 - 3x + x - \sqrt{3} = 0 \)
\( \sqrt{3}x(x - \sqrt{3}) + 1(x - \sqrt{3}) = 0 \)
\( (x - \sqrt{3})(\sqrt{3}x + 1) = 0 \)
\( \therefore x = \sqrt{3} \) or \( -\frac{1}{\sqrt{3}} \).

Question. For what values of \( k \), the given quadratic equation \( 9x^2 + 6kx + 4 = 0 \) has equal roots?
Answer: For equal roots \( D = 0 \)
\( b^2 - 4ac = 0 \)
\( (6k)^2 - 4 \times 9 \times 4 = 0 \)
\( 36k^2 = 144 \)
\( k^2 = 4 \)
\( k = \pm 2 \).

Question. For what value(s) of 'a' quadratic equation \( 3ax^2 - 6x + 1 = 0 \) has no real roots?
Answer: Given that, \( 3ax^2 - 6x + 1 = 0 \)
For no real roots, \( D < 0 \)
Discriminant, \( D < 0 \)
\( (-6)^2 - 4(3a)(1) < 0 \)
\( 36 - 12a < 0 \)
\( 12a > 36 \)
\( a > 3 \).

Question. Find the values(s) of \( k \) for which the quadratic equation \( x^2 + 2\sqrt{2}kx + 18 = 0 \) has equal roots.
Answer: Since, \( x^2 + 2\sqrt{2}kx + 18 = 0 \) has equal roots
\( D = 0 \)
\( b^2 = 4ac \)
\( (2\sqrt{2}k)^2 = 4 \times 1 \times 18 \)
\( 4 \times 2 \times k^2 = 72 \)
\( 8 \times k^2 = 72 \)
\( k^2 = \frac{72}{8} \)
\( k^2 = 9 \)
\( k = \pm 3 \).

Question. For what values of \( k \), the roots of the equation \( x^2 + 4x + k = 0 \) are real?
Answer: Since roots of the equation \( x^2 + 4x + k = 0 \) are real
\( \Rightarrow D \ge 0 \)
\( \Rightarrow b^2 - 4ac \ge 0 \)
\( \Rightarrow 4^2 - 4 \times 1 \times k \ge 0 \)
\( \Rightarrow 16 - 4k \ge 0 \)
\( \Rightarrow 4k \le 16 \)
\( \Rightarrow k \le 4 \).

Question. Find the nature of roots of the quadratic equation \( 2x^2 - 4x + 3 = 0 \).
Answer: \( 2x^2 - 4x + 3 = 0 \)
\( D = b^2 - 4ac = (-4)^2 - 4(2)(3) \)
\( D = 16 - 24 = -8 \)
\( \therefore \) Equation has no real roots.